Priority of the logical operators (order of operations) for NOT, AND, OR in Python

As far as I know, in C & C++, the priority sequence for NOT AND & OR is NOT>AND>OR. But this doesn't seem to work in a similar way in Python. I tried searching for it in the Python documentation and failed (Guess I'm a little impatient.). Can someone clear this up for me?

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not binds tighter than and which binds tighter than or as stated in the language reference

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It's NOT, AND, OR, from highest to lowest according to the documentation on Operator precedence

Here is the complete precedence table, lowest precedence to highest. A row has the same precedence and groups from left to right

 0. :=
1. lambda
2. if – else
3. or
4. and
5. not x
6. in, not in, is, is not, <, <=, >, >=, !=, ==
7. |
8. ^
9. &
10. <<, >>
11. +, -
12. *, @, /, //, %
13. +x, -x, ~x
14. **
14. await x
15. x[index], x[index:index], x(arguments...), x.attribute
16. (expressions...), [expressions...], {key: value...}, {expressions...}
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Of the boolean operators the precedence, from weakest to strongest, is as follows:

  1. or
  2. and
  3. not x
  4. is not; not in

Where operators are of equal precedence evaluation proceeds from left to right.

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There is no good reason for Python to have other priority sequence of those operators than well established one in (almost) all other programming languages, including C/C++.

You may find it in The Python Language Reference, part 6.16 - Operator precedence, downloadable (for the current version and packed with all other standard documentation) from https://docs.python.org/3/download.html, or read it online here: 6.16. Operator precedence.

But there is still something in Python which can mislead you: The result of and and or operators may be different from True or False - see 6.11 Boolean operations in the same document.

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You can do the following test to figure out the precedence of and and or.

First, try 0 and 0 or 1 in python console

If or binds first, then we would expect 0 as output.

In my console, 1 is the output. It means and either binds first or equal to or (maybe expressions are evaluated from left to right).

Then try 1 or 0 and 0.

If or and and bind equally with the built-in left to right evaluation order, then we should get 0 as output.

In my console, 1 is the output. Then we can conclude that and has higher priority than or.

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Some simple examples; note the operator precedence (not, and, or); parenthesize to assist human-interpretability.

a = 'apple'
b = 'banana'
c = 'carrots'


if c == 'carrots' and a == 'apple' and b == 'BELGIUM':
print('True')
else:
print('False')
# False

Similarly:

if b == 'banana'
True


if c == 'CANADA' and a == 'apple'
False


if c == 'CANADA' or a == 'apple'
True


if c == 'carrots' and a == 'apple' or b == 'BELGIUM'
True


# Note this one, which might surprise you:
if c == 'CANADA' and a == 'apple' or b == 'banana'
True


# ... it is the same as:
if (c == 'CANADA' and a == 'apple') or b == 'banana':
True


if c == 'CANADA' and (a == 'apple' or b == 'banana'):
False


if c == 'CANADA' and a == 'apple' or b == 'BELGIUM'
False


if c == 'CANADA' or a == 'apple' and b == 'banana'
True


if c == 'CANADA' or (a == 'apple' and b == 'banana')
True


if (c == 'carrots' and a == 'apple') or b == 'BELGIUM'
True


if c == 'carrots' and (a == 'apple' or b == 'BELGIUM')
True


if a == 'apple' and b == 'banana' or c == 'CANADA'
True


if (a == 'apple' and b == 'banana') or c == 'CANADA'
True


if a == 'apple' and (b == 'banana' or c == 'CANADA')
True


if a == 'apple' and (b == 'banana' and c == 'CANADA')
False


if a == 'apple' or (b == 'banana' and c == 'CANADA')
True
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Expression 1 or 1 and 0 or 0 returns 1. Looks like we have the same priority, almost same.

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not > and

print(~0&0) # 0

and > or

print(0&0|1) # 1

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